Let $U \subset \Bbb{R}^n$ be an open bounded smooth domain, let $u \in H^1_0(U)$ satisfy the Poisson equation:
$$-\Delta u = f \tag{*}$$
with $f \in L^{2}(U)$, by the classical elliptic regularity result we have the following estimate see for example here :
$$\|u\|_{H^2(U)} \le C(\|u\|_{L^2} + \|f\|_{L^2}) \tag{i}$$
Correct? However I see in some place the regularity estimate is written as follows see here:
$$\|u\|_{H^2(U)} \le C\|f\|_{L^2(U)} \tag{ii}$$
Which is a better estimate than the first one. How can I get (ii)?
More generally, Let $Lu = \sum_{i,j} (a_{ij} u_{x_i})_{x_j}$ be the elliptic operator. Then we have $$\int a_{ij} u_{x_i}u_{x_j} = (f,u)$$
Therefore by ellipticity:
$$\theta \|Du\|_{L^2(U)}^2 \le \|f\|_{L^2}\|u\|_{L^2}$$
Combine with Poincare inequality we have $$\theta \|Du\|^{2}_{L^2(U)} \le C\|f\|_{L^2}\|Du\|_{L^2}$$
Therefore $$\|Du\|_{L^2} \le C\|f\|_{L^2}$$ Combine with Pincare inequality again we have $$\|u\|_{L^2} \le C\|f\|_{L^2} \tag{**}$$
Therefore combine with (i) above we get (ii).
However when $L$ has zero or first order term the last estimate (**) in this answer does not holds.